Answer:
P(X>17) = 0.979
Step-by-step explanation:
Probability that a camera is defective, p = 3% = 3/100 = 0.03
20 cameras were randomly selected.i.e sample size, n = 20
Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97
Probability that more than 17 cameras are working P ( X > 17)
This is a binomial distribution P(X = r) [tex]nCr q^{r} p^{n-r}[/tex]
[tex]nCr = \frac{n!}{(n-r)!r!}[/tex]
P(X>17) = P(X=18) + P(X=19) + P(X=20)
P(X=18) = [tex]20C18 * 0.97^{18} * 0.03^{20-18}[/tex]
P(X=18) = [tex]20C18 * 0.97^{18} * 0.03^{2}[/tex]
P(X=18) = 0.0988
P(X=19) = [tex]20C19 * 0.97^{19} * 0.03^{20-19}[/tex]
P(X=19) = [tex]20C19 * 0.97^{19} * 0.03^{1}[/tex]
P(X=19) = 0.3364
P(X=20) = [tex]20C20 * 0.97^{20} * 0.03^{20-20}[/tex]
P(X=20) = [tex]20C20 * 0.97^{20} * 0.03^{0}[/tex]
P(X=20) = 0.5438
P(X>17) = 0.0988 + 0.3364 + 0.5438
P(X>17) = 0.979
The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch
